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Question on Dirac Delta and functional derivatives

  1. Jan 25, 2007 #1
    Suppose that we have a compact manifold [itex]\mathcal{M}[/itex] with a positive definite metric [itex]g_{ij}[/itex]. The volume of the manifold is then

    [tex]
    V = \int_\mathcal{M} d^3x \sqrt{g(x)},
    [/tex]

    where [itex]x^i[/itex] are coordinates on [itex]\mathcal{M}[/itex] and [itex]\sqrt{g(x)}[/itex] is the square root of the determinant of the metric. Suppose now that we take a functional derivative of the volume with respect to the metric, i.e., we want to calculate

    [tex]
    \frac{\delta}{\delta g_{ij}(y)} V
    = \frac{\delta}{\delta g_{ij}(y)}\int_\mathcal{M} d^3x \sqrt{g(x)}
    [/tex]

    As far as I know, this calculation gives

    [tex]
    \frac{\delta}{\delta g_{ij}(y)} V
    = \frac{1}{2}\int_\mathcal{M} d^3x \sqrt{g(x)} g^{ij}(x) \delta^{(3)}(x,y)
    [/tex]

    Where [itex]\delta^{(3)}(x)[/itex] is a three-dimensional Dirac delta function. The question I have is this: is [itex]\delta^{(3)}(x,y)[/itex] to be regarded as (a) a unit weight density or (b) a density of weight zero?

    If the answer is (a) then the result is

    [tex]
    \frac{\delta}{\delta g_{ij}(y)} V
    = \frac{1}{2}\sqrt{g(y)} g^{ij}(y)
    [/tex]

    while if the answer is (b) then the result for the functional derivative is

    [tex]
    \frac{\delta}{\delta g_{ij}(y)} V
    = \frac{1}{2} g^{ij}(y)
    [/tex]

    So which one is the correct answer?
     
    Last edited: Jan 25, 2007
  2. jcsd
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